MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Edit according to the comment of user36931 I remove the "motivation" from the previous version and I add an statement to the first question

We consider the following two classes of smooth maps on $\mathbb{R}^{n}.$ ( $n$ is not necessarily even):

$HP_{1}$: $\;$A smooth map $f:\mathbb{R}^{n} \to \mathbb{R^{n}}$ satisfies property $HP_{1}$ if the pull back metric $f^{*}(g)$ is a conformal metric for every conformal metric $g$ on $\mathbb{R}^{n}.$ ( By conformality I mean $g=e^{h} \sum dx_{i}^{2})$

$HP_{2}$: $\;$A smooth map $X$ on $\mathbb{R}^{n}$ with coordinates $X=(P_{1},P_{2},\ldots,P_{n})$ satisfies $HP_{2}$ if the space of harmonic functions is invariant under the derivational operator $U \mapsto X.U=\sum P_{i} \partial U/ \partial x_{i}$


  1. Assume $n=2k$, identify $\mathbb{R}^{2k}$ with $\mathbb{C}^{k}$. Let $f$ satisfies $HP_{1}$. Is $f$ a holomorphic map on $\mathbb{C}^{k}$? The same question for $HP_{2}$?

  2. Are the above two properties, equivalents?(A map is $HP_{1}$ iff it is $HP_{2}$?)

share|cite|improve this question
Why does your property hold for maps C^2 \to C^2 ? Also see…;. – user36931 Apr 6 '14 at 13:14
By the above i meant property one. I see it if the map is "diagonal" but not if it mixes up the two factors. – user36931 Apr 6 '14 at 13:26
@user36931 However I wrote in the "motivation" "HP_{2} implies holomorphicity", but now I realize that my argument was not complete! What is your reason for "diagonal maps"?(Note that HP2 was that "harmonic maps are invariant under derivation of X" th question is X holomorphic? – Ali Taghavi Apr 7 '14 at 4:13
up vote 7 down vote accepted

This is an extended comment. The first property $HP_1$ simply says that $f$ is conformal. It is not true that holomorphic maps of $C^n=R^{2n}$ are conformal, except when $n=1$. In fact there are very few conformal maps in $R^n$ for $n\geq 3$: they are only Mobius transfomrations (compositions of reflections).

share|cite|improve this answer
Moreover conformal maps need not be holomorphic (even rotations can send a complex line to a totally real plane). – Benoît Kloeckner Apr 6 '14 at 18:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.